I'm given two exponential random variables $x_1$, $x_2$ that both have a mean of $1$ (so $\lambda = 1$ for both r.v.s, presumably), and asked to solve for the probability: $$P(X_{(2)} \gt 3*X_{(1)})$$
Not sure about how to approach this. I think I've solved for the joint distribution of the order statistics correctly ($f_{x_{(1)},x_{(2)}}(x_1,x_2)=2e^{-x_1}*e^{-x_2}$ when $0 \lt x_1\lt x_2$), but don't know where to go for there.
Thanks for y'alls help, in advance.
The joint density function is correct. For the required probability, draw a picture, calling the usual $x$-axis the $x_1$-axis, and the usual $y$-axis the $x_2$-axis. Find $$\iint_A 2e^{-x_1}e^{-x_2}\,dx_1\,dx_2,$$ where $A$ is the part of the first quadrant that is above the line $x_2=3x_1$. This can be expressed as the iterated integral $$\int_{0}^\infty \left(\int_{3x_1}^\infty 2e^{-x_1}e^{-x_2}\,dx_2\right)\,dx_1.$$
Remark: In this case, we do not need the joint density function of the order statistics. Let our two exponential random variables be $X$ and $Y$. With probability $\frac{1}{2}$, we have $X\lt Y$, and with probability $\frac{1}{2}$ we have $Y\lt X$. Thus the required probability is $$\frac{1}{2}\Pr(Y\gt 3X)+\frac{1}{2}\Pr(X\gt 3Y).$$ Each probability is computed by finding a double integral very much in the style of the above answer.